Probability

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Applicable MathematicsApplicable Mathematics
“Probability”“Probability”

DefinitionsDefinitions
Probability is the mathematics of chance.
It tells us the relative frequency with which we can
expect an event to occur
The greater the probability the more
likely the event will occur.
It can be written as a fraction, decimal, percent,
or ratio.

Certain
Impossible
.5
1
0
50/50
Probability is the numerical
measure of the likelihood that
the event will occur.
Value is between 0 and 1.
Sum of the probabilities of all events
is 1.

A probability experiment is an action through which
specific results (counts, measurements, or responses)
are obtained.
The result of a single trial in a probability experiment
is an outcome.
The set of all possible outcomes of a probability
experiment is the sample space, denoted as S.
e.g. All 6 faces of a die: S = { 1 , 2 , 3 , 4 , 5 , 6 }

Other Examples of Sample Spaces may include:
Lists
Tables
Grids
Venn Diagrams
Tree Diagrams
May use a combination of these
Where appropriate always display your sample space

An event consists of one or more outcomes and is a
subset of the sample space.
Events are often represented by uppercase letters,
such as A, B, or C.
Notation: The probability that event E will occur is
written P(E) and is read
“the probability of event E.”

• The Probability of an Event, E:
Consider a pair of Dice
• Each of the Outcomes in the Sample Space are random
and equally likely to occur.e.g. P( ) =
(There are 2 ways to get one 6 and the other 4)
P(E) =
Number of Event Outcomes
Total Number of Possible Outcomes in S
18
1
36
2
=

There are three types of probability
1. Theoretical Probability
Theoretical probability is used when each outcome
in a sample space is equally likely to occur.
P(E) =
Number of Event Outcomes
Total Number of Possible Outcomes in S
The Ultimate probability formula 

2. Experimental Probability
Experimental probability is based upon observations
obtained from probability experiments.
The experimental probability of an event
E is the relative frequency of event E
P(E) =
Number of Event Occurrences
Total Number of Observations

3. Subjective Probability
Subjective probability is a probability measure
resulting from intuition, educated guesses, and
estimates.
Therefore, there is no formula to calculate it.
Usually found by consulting an expert.

Law of Large Numbers.
As an experiment is repeated over and over, the
experimental probability of an event approaches
the theoretical probability of the event.
The greater the number of trials the more likely
the experimental probability of an event will equal
its theoretical probability.

Complimentary EventsComplimentary Events
The complement of event E is the set of all
outcomes in a sample space that are not included in
event E.
The complement of event E is denoted by
Properties of Probability:
EorE′
)(1)(
)(1)(
1)()(
1)(0
EPEP
EPEP
EPEP
EP
−=
−=
=+
≤≤

The Multiplication RuleThe Multiplication Rule
If events A and B are independent, then the
probability of two events, A and B occurring in a
sequence (or simultaneously) is:
This rule can extend to any number of independent
events.
)()()()( BPAPBAPBandAP ×=∩=
Two events are independent if the occurrence of the
first event does not affect the probability of the
occurrence of the second event. More on this later

Mutually ExclusiveMutually Exclusive
Two events A and B are mutually exclusive if and
only if:
In a Venn diagram this means that event A is
disjoint from event B.
A and B are M.E.
0)( =∩ BAP
A BA B
A and B are not M.E.

The Addition RuleThe Addition Rule
The probability that at least one of the events A
or B will occur, P(A or B), is given by:
If events A and B are mutually exclusive, then the
addition rule is simplified to:
This simplified rule can be extended to any number
of mutually exclusive events.
)()()()()( BAPBPAPBAPBorAP ∩−+=∪=
)()()()( BPAPBAPBorAP +=∪=

Conditional ProbabilityConditional Probability
Conditional probability is the probability of an event
occurring, given that another event has already
occurred.
Conditional probability restricts the sample space.
The conditional probability of event B occurring,
given that event A has occurred, is denoted by P(B|
A) and is read as “probability of B, given A.”
We use conditional probability when two events
occurring in sequence are not independent. In other
words, the fact that the first event (event A) has
occurred affects the probability that the second
event (event B) will occur.

Formula for Conditional Probability
Better off to use your brain and work out
conditional probabilities from looking at the sample
space, otherwise use the formula.
)(
)(
)|(
)(
)(
)|(
AP
ABP
ABPor
BP
BAP
BAP
∩
=
∩
=

Conditional probability problems can be solved by
considering the individual possibilities or by using a
table, a Venn diagram, a tree diagram or a formula.
Harder problems are best solved by using a formula
together with a tree diagram.
e.g. There are 2 red and 3 blue counters in a bag
and, without looking, we take out one counter and
do not replace it.
The probability of a 2nd
counter taken from the bag
being red depends on whether the 1st
was red or
blue.

e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
One person is selected at random.
42123Female
73312Male
HighMediumLow
100
Total
L is the event “the person owns a low rated car”

Female
Low
Male
42123
73312
HighMedium
F is the event “a female is chosen”.
100
Total

Female
Low
Male
42123
73312
HighMedium
F is the event “a female is chosen”.
There is no need for a Venn diagram or a formula to
solve this type of problem.
We just need to be careful which row or column we look at.
Find (i) P(L) (ii) P(F ∩ L) (iii) P(F L)
100
Total

23
12
100100
(i) P(L) =
Solution:
35
421Female
733Male
TotalHighMediumLow
(Best to leave the answers as fractions)
20 20
7
=
23
12
735
100
Low

(i) P(L) =
Solution:
100
42123Female
73312Male
HighMediumLow
20 20
7
=
735
100
(ii) P(F ∩ L) = The probability of selecting a
female with a low rated car.
23
23
100
Total
100

(i) P(L) =
Solution:
10035
42123Female
73312Male
HighMediumLow
20 20
7
=
735
100
(ii) P(F ∩ L) =
23
Total
100
(iii) P(F L) = The probability of selecting a
female given the car is low rated.
23 We must be careful with the
denominators in (ii) and (iii). Here
we are given the car is low rated.
We want the total of that column.
23
35
12
The sample space is restricted
from 100 to 35.

(i) P(L) =
Solution:
100
42123Female
73312Male
HighMediumLow
20 20
7
=
735
100
(ii) P(F ∩ L) =
23
Total
100
(iii) P(F L) =
23
Notice that
P(L) × P(F L)
35
23
20
7
×=
100
23
=
So, P(F ∩ L) = P(F|L) × P(L)
= P(F ∩ L)
35
5
1

e.g. 2. I have 2 packets of seeds. One contains 20
seeds and although they look the same, 8 will give red
flowers and 12 blue. The 2nd packet has 25 seeds of
which 15 will be red and 10 blue.
Draw a Venn diagram and use it to illustrate the
conditional probability formula.
Solution: Let R be the event “ Red flower ” and
F be the event “ First packet ”
RF
Red in the 1st
packet

Solution:
RF
8
Red in the 1st
packet
Let R be the event “ Red flower ” and

Solution:
RF
Blue in the 1st
packet
8

Solution:
RF
Blue in the 1st
packet
812

Solution:
RF
Red in the 2nd
packet
812

Solution:
RF
Red in the 2nd
packet
15812

Solution:
RF
15812
Blue in the 2nd
packet

Solution:
RF
15
10
812
Blue in the 2nd
packet

Solution:
RF
15
10
812
Total: 20 + 25

45
Solution:
RF
15
10
812
Total: 20 + 25

45
Solution:
RF
15
10
812

45
RF
Solution:
15
10
812
P(R ∩ F) =

8 45
RF
Solution:
15
10
812
P(R ∩ F) =

8 45
RF
Solution:
15
10
812
P(R ∩ F) =
45

45
RF
Solution:
15
10
12
P(R ∩ F) =
P(R F) =
8
8
45
8

45
RF
Solution:
15
10
12
P(R ∩ F) =
P(F) =P(R F) =
8
8
45
20
8

45
RF
Solution:
15
10
12
P(R ∩ F) =
P(F) =
20
P(R F) =
8
8
45
20
8

45
RF
Solution:
15
10
12
P(R ∩ F) =
P(F) =
20
P(R F) =
8
8
45
20
8
45

45
RF
Solution:
15
10
12
P(R ∩ F) =
P(F) =P(R F) =
8
8
45
20
8
45
20
20
8
×
P(R ∩ F) = P(R|F) × P(F)So,
P(R F) × P(F) =⇒
20
45
45
8
=
1
1

Probability TreeProbability Tree
DiagramsDiagrams
The probability of a complex event can be found
using a probability tree diagram.
1. Draw the appropriate tree diagram.
2. Assign probabilities to each branch.
(Each section sums to 1.)
3. Multiply the probabilities along individual branches
to find the probability of the outcome at the end of
each branch.
4. Add the probabilities of the relevant outcomes,
depending on the event.

e.g. 3. In November, the probability of a man getting
to work on time if there is fog on the M6 is .5
2
10
9
If the visibility is good, the probability is .
20
3
The probability of fog at the time he travels is .
(a) Calculate the probability of him arriving on time.
There are lots of clues in the question to tell us we are
dealing with conditional probability.
(b) Calculate the probability that there was fog given that
he arrives on time.

e.g. 3. In November, the probability of a man getting
to work on time if there is fog on the M6 is .5
2
10
9
20
3
If the visibility is good, the probability is .
The probability of fog at the time he travels is .
he arrives on time.
There are lots of clues in the question to tell us we are
dealing with conditional probability.
Solution: Let T be the event “ getting to work on time ”
Let F be the event “ fog on the M6 ”
Can you write down the notation for the probabilities
that we want to find in (a) and (b)?

“ the probability of a man getting to work on time if
there is fog is ”5
2
10
9
“ If the visibility is good, the probability is ”.
20
3
“ The probability of fog at the time he travels is ”.
Can you also write down the notation for the three
probabilities given in the question?
This is a much harder problem so we draw a tree diagram.
P(F T)
P(T)
5
2
=P(T F)
10
9
=P(T F/
)
20
3
=P(F)
he arrives on time.
Not foggy

Not on
time
5
2
20
17
20
3
10
9
10
1
5
2
20
3
×
5
3
20
3
×
10
9
20
17
×
10
1
20
17
×
Fog
No
Fog
On
time
On
time
Not on
time
5
2
=P(T F)
10
9
=P(T F/
)
20
3
=P(F)
5
3
F
F/
T
T/
T
T/
Each section sums to 1

5
2
=P(T F)
10
9
=P(T F/
)
20
3
=P(F)
5
2
20
17
20
3
10
9
10
1
5
2
20
3
×
5
3
20
3
×
10
9
20
17
×
10
1
20
17
×
5
3
F
F/
T
T/
T
T/
Because we only reach the 2nd
set of branches
after the 1st
set has occurred, the 2nd
set
must represent conditional probabilities.

5
2
20
17
20
3
10
9
10
1
5
2
20
3
×
5
3
20
3
×
10
9
20
17
×
10
1
20
17
×
5
3
F
F/
T
T/
T
T/

5
2
20
17
20
3
10
9
10
1
5
3
20
3
×
10
9
20
17
×
10
1
20
17
×
5
3
F
F/
T
T/
T
T/
5
2
20
3
×
100
6
=
( foggy and he
arrives on time )

5
2
20
17
20
3
10
9
10
1
5
3
20
3
×
5
3
F
F/
T
T/
T
T/
10
9
20
17
×
10
1
20
17
×
5
2
20
3
×
100
6
=
200
153
=
=+=
200
153
100
6
200
165
)()()( T
/
FTFT ∩∩ += PPP
33
40 40
33
=
( not foggy and he
arrives on time )

Fog on M 6 Getting to work
F
T
20
3
5
2
40
33
)( =TPFrom part (a),
=∩ )( TFP
100
6
5
2
20
3
=×
he arrives on time.
We need )( TFP
)(
)(
)(
T
TF
T|F
P
P
P
∩
=
55
4
( =⇒ )P TF
)(
)(
(
T
TF
T)|F
P
P
P
∩
=
⇒
40
33
100
6
( ÷=)TFP
33
40
100
6
×=
5
22
11

Eg 4. The probability of a maximum temperature of 28° or
more on the 1st
day of Wimbledon ( tennis competition! )
8
3
4
3
2
1
has been estimated as . The probability of a particular
Aussie player winning on the 1st
day if it is below 28° is
estimated to be but otherwise only .
Draw a tree diagram and use it to help solve the following:
(i) the probability of the player winning,
(ii) the probability that, if the player has won, it was at
least 28°.
Solution: Let T be the event “ temperature 28° or more ”
Let W be the event “ player wins ”
8
3
)( =TPThen,
4
3
)( =/
TWP
2
1
)( =TWP

8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
High
temp
W
Wins
Loses
Loses
Lower
temp
Sum
= 1
Let T be the event “ temperature 28° or more ”
Let W be the event “ player wins ”
8
3
)( =TPThen,
4
3
)( =/
TWP
2
1
)( =TWP
T
T/
Wins
W
W/
W/

)()()( W
/
TWTW ∩∩ += PPP(i)
8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
W
T
T/
W
W/
W/

32
21
32
156
32
15
16
3
=
+
=+=)()()( W
/
TWTW ∩∩ += PPP(i)
8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
W
T
T/
W
W/
W/

32
21
)( =WP
8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
W
T
T/
W
W/
W/
)(
)(
)(
W
WT
W|T
P
P
P
∩
=(ii)

32
21
)( =WP
)(
)(
)(
W
WT
W|T
P
P
P
∩
=(ii)
32
21
16
3
)( ÷=⇒ WTP
8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
W
T
T/
W
W/
W/

7
2
=
32
21
)( =WP
)(
)(
)(
W
WT
W|T
P
P
P
∩
=(ii)
21
32
16
3
×=
8
3
2
1
2
1
4
3
4
1
16
3
2
1
8
3
=×
16
3
2
1
8
3
=×
32
15
4
3
8
5
=×
32
5
4
1
8
5
=×
8
5
W
T
T/
W
W/
W/
7
1 2
132
21
16
3
)( ÷=⇒ WTP

We can deduce an important result from the conditional
law of probability:
( The probability of A does not depend on B. )
or P(A ∩ B) = P(A) × P(B)
If B has no effect on A, then, P(A B) = P(A) and we say
the events are independent.
becomes P(A) =
P(A ∩ B)
P(B)
P(A|B) =So, P(A ∩ B)
P(B)
Independent EventsIndependent Events

Tests for independence
P(A ∩ B) = P(A) × P(B)
P(A B) = P(A)
or
Independent EventsIndependent Events
P(B A) = P(B)

∑=
=
n
i
ii pxxE
1
)(
Expected ValueExpected Value
Suppose that the outcomes of an experiment are
real numbers called
and suppose that these outcomes have probabilities
respectively. Then the expected
nxxxx ,...,,, 321
npppp ,...,,, 321
value of x, E(x), of the experiment
is:

Expected ValueExpected Value
Example
At a raffle, 1500 tickets are sold at $2 each for
four prizes of $500, $250, $150, and $75. What
is the expected value of your gain if you play?
P(x)
Gain -$2$73$148$248$498
1500
1
1500
1
1500
1
1500
1
1500
1496
35.1$
1500
1496
2
1500
1
73
1500
1
148
1500
1
248
1500
1
498)(
−=
×−+×+×+×+×=xE

OddsOdds
When one speaks about the odds in favour of an
event, they are actually stating the number of
favourable outcomes of an event to the number
of unfavourable outcomes of the event, assuming
that the outcomes are equally likely.
The odds in favour of event E are:
n(E):n(E) P(E):P(E)
The odds against event E are
n(E):n(E) P(E):P(E)
If the odds in favour of E are a:b, then
ba
a
EP
+
=)(

Probability

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